[Back to TUTOR SWAG index] [Back to Main SWAG index] [Original]
A introduction to x86 assembly
Written by Gavin Estey email: gavin@senator.demon.co.uk
This document is version:
0.9 Beta 2/19/95
This will hopefully be the final version of this unless somebody finds some
mistakes in it. I have spent over 620 minutes creating this document (not
including writing or testing the code or the original text document) so I hope
you will find it useful.
This version is still in testing which means that I cannot be sure that all the code
will compile and work correctly. I have done my best to make sure that all the
information contained in this document is correct. If any mistakes are found
please could you notify me be email.
If someone who has a computer that doesn�t have a VGA card/monitor could you
please compile the code on P29 that checks what graphics the computer can handle
and tell me if it returned DX as 0.
This document was written by Gavin Estey. No part of this can be reproduced and sold
in any commercial product without my written permission (email me for more
information). I am not responsible for any damage caused in anyway from this
document.
If you enjoy this document then a donation would be gratefully received. No payment
is necessary but none will turned down. If any mistakes are noticed then I would like to
be notified. If you would like any other help or have suggestions for a later version of
this document.
Firstly you will need a suitable assembler to compile your programs. All examples have
been tested with two compilers: A86 and Turbo Assemblerâ ver 4. A86 is a very
good shareware assembler capable of producing code up to 80286. This can be found
on your local simtel mirror (try ftp.demon.co.uk or ftp.cdrom.com) under
simtel/msdos/asmutil called A86V322.ZIP.
Firstly I am going to talk about SEGMENTS and OFFSETS. This (for me anyway) is
probably the hardest part of assembly to understand.
Segments and offsets:
The original designers of the 8088 decided that nobody will every need to use more
that one megabyte of memory. So they built the chip so it couldn't access above that.
The problem is to access a whole megabyte 20 bits are needed (one bit being either a
one or a zero). Registers only have 16 bits and they didn't want to use two because
that would be 32 bits and they thought that this would be too much for anyone. They
decided to do the addressing with two registers but not 32 bits. Is this confusing?
Blame the designers of the 8088.
OFFSET = SEGMENT * 16
SEGMENT = OFFSET / 16 (the lower 4 bits are lost)
SEGMENT * 16 |0010010000010000 - ---| range (0 to 65535) * 16
OFFSET |-- - -0100100000100010| range (0 to 65535)
20 bit address |00101000100100100010| range 0 to 1048575 (1 MEG)
===== DS =====
====== SI =====
(note DS and SI overlap). This is how DS:SI is used to make a 20 bit address. The
segment is in DS and the offset is in SI.
Segment registers are: CS, DS, ES, SS. On the 386+ there are also FS & GS
Offset registers are: BX, DI, SI, BP, SP, IP. In 386+ protected mode, ANY general
register (not a segment register) can be used as an Offset register. (Except IP, which
you can't access.)
Registers:
AX, BX, CX and DX are general purpose registers. On a 386+ they are 32 bits; EAX,
EBX, ECX and EDX. The all can be split up into high and low parts, each being 8 bits.
EAX 32 bits
AX 16 bits
AH 8 bits
AL 8 bits
This means that you can use AH and AL to store different numbers and treat them as
separate registers for some tasks.
BX (BH/BL): same as AX
CX (CH/CL): Same as AX (used for loops)
DX (DH/DL): Same as AX (used for multiplication/division)
DI and SI are index registers and can be used as offset registers.
SP: Stack Pointer. Does just that. Points to the current position in the stack. Don't
alter unless you REALLY know what you are doing or want to crash your computer.
The Stack:
This is and area of memory that is like a stack of plates. The last one you put on is the
first one that you take off (LOFO). If another piece of data is put on the stack it grows
downwards.
Diagram 1: This shows how the stack is organised
PUSH and POP:
Push puts a value onto the stack and pop takes it back off. Here's some code. (you
can't compile this - yet.)
push cx ;put cx on the stack
push ax ;put ax on the stack
.
.
. ;this is going to pop them back in
. ;the wrong so they are reversed
pop cx ;put value from stack into cx
pop ax ;put value from stack into ax
This means that cx is equal to what ax is and ax is equal to what cx was.
Some instructions:
MOV: this moves a value from one place to another.
Syntax:
MOV destination, source
for example:
MOV ax,10 ;moves an immediate value into ax
MOV bx,cx ;moves value from cx into bx
INT: calls a DOS or BIOS function, mainly to do something you would rather not
write a function for e.g. change video mode, open a file etc.
Syntax:
INT interrupt number
Note: an h after a number denotes that that number is in hex.
INT 21h ;Calls DOS standard interrupt 21h
INT 10h ;Calls the Video BIOS interrupt
All interrupts need a value to specify what subroutine to use. This is usually in AH. To
print a message on the screen all you need to do is this:
MOV ah,9 ;subroutine number 9
INT 21h ;call the interrupt
But first you have to specify what to print. This function needs DS:DX to be a far
pointer to where the string is. The string has to be terminated with a dollar sign ($).
This example shows how it works:
MyMessage db "This is a message!$"
.
.
.
mov dx,OFFSET MyMessage
mov ax,SEG MyMessage
mov ds,ax
mov ah,9
int 21h
DB declares a piece of data. This is the same as in BASIC:
LET MyMessage$ = "This is a message!"
DB declares a byte, DW declares a word and DD declares a Dword.
This is your first assembly program. Cut this in and then assemble it. If you are using
A86, type:
A86 name of file
If you are using Turbo Assembler then type:
tasm name of file
tlink name of file minus extention
With tlink you can make a .COM file by using a /t switch.
;THIS IS A SIMPLE PROGRAM TO DISPLAY A MESSAGE ON THE
;SCREEN. SHOULD OUTPUT This is a message! TO SCREEN AND
;RETURN WITH NO ERRORLEVEL. THIS CAN�T BE MADE INTO A
.COM FILE WITH TLINK AS IT GENERATES AN ERROR.
.MODEL SMALL ;Ignore the first two lines
;I'll explain what this means later.
.STACK ;allocate a stack
.CODE ;a directive to start code segment
START: ;generally a good name to use as an
;entry point
JMP BEGIN ;jump to BEGIN: (like goto)
MyMessage db "This is a message!$"
BEGIN: ;code starts here
mov dx,OFFSET MyMessage ;put the offset of message
;in DX
mov ax,SEG MyMessage ;put the segment that the
;message is in in AX
mov ds,ax ;copy AX into DS
mov ah,9 ;move 9 into ah to call
int 21h ;function 9 of interrupt
;21h - display a string to
;screen.
MOV ax,4c00h ;move the value of 4c00 hex into AX
INT 21h ;call interrupt 21h, subroutine 4C
;which returns control to dos, otherwise it would of
;crashed. The 00 on the end means what errorlevel it
;would return to DOS. This can be checked in a batch file
END START ;end here
Some instructions that you need to know:
This is just a list of some basic assembly instructions that are very important and are
used often.
ADD Add the contents of one number to another
Syntax:
ADD operand1,operand2
This adds operand2 to operand1. The answer is stored in operand1. Immediate data
cannot be used as operand1 but can be used as operand2.
SUB Subtract one number from another
Syntax:
SUB operand1,operand2
This subtracts operand2 from operand1. Immediate data cannot be used as operand1
but can be used as operand2.
MUL Multiplies two unsigned integers (always positive)
IMUL Multiplies two signed integers (either positive or negitive)
Syntax:
MUL register or variable
IMUL register or variable
This multiples the register given by the number in AL or AX depending on the size of
the operand. The answer is given in AX. If the answer is bigger than 16 bits then the
answer is in DX:AX (the high 16 bits in DX and the low 16 bits in AX).
On a 386, 486 or Pentium the EAX register can be used and the answer is stored in
EDX:EAX.
DIV Divides two unsigned integers(always positive)
IDIV Divides two signed integers (either positive or negitive)
Syntax:
DIV register or variable
IDIV register or variable
This works in the same way as MUL and IMUL by dividing the number in AX by the
register or variable given. The answer is stored in two places. AL stores the answer
and the remainder is in AH. If the operand is a 16 bit register than the number in
DX:AX is divided by the operand and the answer is stored in AX and remainder in
DX.
The way we entered the address of the message we wanted to print was a bit
cumbersome. It took three lines and it isn�t the easiest thing to remember
mov dx,OFFSET MyMessage
mov ax,SEG MyMessage
mov ds,ax
We can replace all this with just one line. This makes the code easier to read and it
easier to remember. This only works if the data is only in the data is in one segment i.e.
small memory model.
lea dx,MyMessage
or mov dx,OFFSET MyMessage
Using lea is slightly slower and results in code which is larger.
LEA means Load Effective Address.
Syntax:
LEA destination,source
Desination can be any 16 bit register and the source must be a memory operand (bit of
data in memory). It puts the offset address of the source in the destination.
Keyboard input:
We are going to use interrupt 16h, function 00h to read the keyboard. This gets a key
from the keyboard buffer. If there isn't one, it waits until there is. It returns the SCAN
code in AH and the ASCII translation in AL.
MOV ah,00h ;function 00h of
INT 16h ;interrupt 16h
All we need to worry about for now is the ascii value which is in al.
Printing a character:
The problem is that we have the key that has been pressed in ah. How do we display
it? We can't use function 9h because for that we need to have already defined the string
which has to end with a dollar sign. This is what we do instead:
;after calling function 00h of interrupt 16h
MOV dl,al ;move al (ascii code) into dl
MOV ah,02h ;function 02h of interrupt 21h
INT 21h ;call interrupt 21h
If you want to save the value of ah then push it before and pop it afterwards.
Control flow:
Firstly, the most basic command:
JMP label
This is the same as GOTO in basic.
JMP ALabel
.
.
ALabel:
;code to do something
What do we do if we want to compare somthing. We have just got a key from the user
but we want to do something with it. Lets print something out if it is equal to
somethine else. How do we do that? Its easy. We use the jump on condition
commands. Here is a list of them:
Jump on condition commands:
JA
jumps if the first number was above the second number
JAE
same as above, but will also jump if they are equal
JB
jumps if the first number was below the second
JBE
same as above, but will also jump if they are equal
JNA
jumps if the first number was NOT above (same as JBE)
JNAE
jumps if the first number was NOT above or the same as (same as JB)
JNB
jumps if the first number was NOT below (same as JAE)
JNBE
jumps if the first number was NOT below or the same as (same as JA)
JZ
jumps if the two numbers were equal
JE
same as JZ, just a different name
JNZ
jumps if the two numbers are NOT equal
JNE
same as above
[NOTE: There are quite a few more but these are the most useful. If you want the full
list then get a good assembly book]
They are very easy to use.
Syntax:
CMP register containing value, a value
jump command destination
An example of this is:
cmp al,'Y' ;compare the value in al with Y
je ItsYES ;if it is equal then jump to ItsYES
The following program is an example of how to use control and input and output.
.MODEL SMALL
.STACK ;define a stack
.CODE
Start: ;a good place to start.
lea dx,StartUpMessage ;display a message on th e
;screen
mov ah,9 ;using function 09h
int 21h ;of interrupt 21h
lea dx,Instructions ;display a message on the
;screen
mov ah,9 ;using function 09h
int 21h ;of interrupt 21h
mov ah,00h ;function 00h of
int 16h ;interrupt 16h gets a
;character from user
mov bl,al ;save bl
mov dl,al ;move ascii value of key
;pressed to dl
mov ah,02h ;function 02h of
int 21h ;interrupt 21h displays a
;character to screen
cmp bl,'Y' ;is al=Y?
je Thanks ;if yes then goto Thanks
cmp bl,'y' ;is al=y?
je Thanks ;if yes then goto Thanks
jmp TheEnd
Thanks:
lea dx,ThanksMsg ;display message
mov ah,9 ;using function 9
int 21h ;of interrupt 21h
TheEnd:
lea dx,GoodBye ;print goodbye message
mov ah,9 ;using function 9
int 21h ;of interrupt 21h
mov AX,4C00h ;terminate program and
;return to DOS using
INT 21h ;interrupt 21h function 4CH
.DATA
;0Dh,0Ah adds a enter at the beginning
StartUpMessage DB "A Simple Input Program$"
Instructions DB 0Dh,0Ah,"Just press a Y to continue...$�
ThanksMsg DB 0Dh,0Ah,"Thanks for pressing Y!$"
GoodBye DB 0Dh,0Ah,"Have a nice day!$"
END
Procedures:
Assembly, like C and Pascal can have procedures. These are very useful for series of
commands that have to be repeated often.
This is how a procedure is defined:
PROC AProcedure
.
. ;some code to do something
.
RET ;if this is not here then your computer
;will crash
ENDP AProcedure
You can specify how you want the procedure to be called by adding a FAR or a
NEAR after the procedure name. Otherwise it defaults to the memory model you are
using. For now you are better off not doing this until you become more experianced. I
usually add a NEAR in as compilers can�t decide between a near and a far very well.
This means if the jump needs to be far the compiler will warn you and you can change
it.
This is a program which uses a procedure.
;a simple program with a procedure that prints a message
;onto the screen. (Use /t switch with tlink). Should
;display Hello There! on the screen.
.MODEL TINY
.CODE
ORG 100h
MAIN PROC
JMP Start ;skip the data
HI DB "Hello There!$" ;define a message
Start: ;a good place to start.
Call Display_Hi ;Call the procedure
MOV AX,4C00h ;terminate program and return
;to DOS using
INT 21h ;interrupt 21h function 4Ch
Display_Hi PROC ;Defines start of procedure
lea dx,HI ;put offset of message into DX
mov ah,9 ;function 9 of
int 21h ;interrupt 21h
RET ;THIS HAS TO BE HERE
Display_Hi ENDP ;Defines end of procedure
Main ENDP
END MAIN
Memory Models:
We have been using the .MODEL directive to specify what type of memory model we
use, but what does this mean?
Syntax:
.MODEL MemoryModel
Where MemoryModel can be SMALL, COMPACT, MEDIUM, LARGE, HUGE,
TINY OR FLAT.
Tiny:
This means that there is only one segment for both code and data. This type of
program can be a .COM file.
Small:
This means that by default all code is place in one physical segment and likewise all
data declared in the data segment is also placed in one physical segment. This means
that all proedures and variables are addressed as NEAR by pointing at offsets only.
Compact:
This means that by default all elements of code (procedures) are placed in one physical
segment but each element of data can be placed in its own physical segment. This
means that data elements are addressed by pointing at both at the segment and offset
addresses. Code elements (procedures) are NEAR and varaibles are FAR.
Medium:
This is the opposite to compact. Data elements are near and procedures are FAR.
Large:
This means that both procedures and variables are FAR. You have to point at both the
segment and offset addresses.
Flat:
This isn�t used much as it is for 32 bit unsegmented memory space. For this you need a
dos extender. This is what you would have to use if you were writing a program to
interface with a C/C++ program that used a dos extender such as DOS4GW or
PharLap.
Macros:
(All code examples given are for macros in Turbo Assembler. For A86 either see the
documentation or look in the A86 macro example later in this document).
Macros are very useful for doing something that is done often but for which a
procedure can�t be use. Macros are substituted when the program is compiled to the
code which they contain.
This is the syntax for defining a macro:
Name_of_macro macro
;
;a sequence of instructions
;
endm
These two examples are for macros that take away the boring job of pushing and
popping certain registers:
SaveRegs macro
pop ax
pop bx
pop cx
pop dx
endm
RestoreRegs macro
pop dx
pop cx
pop bx
pop ax
endm
Note that the registers are popped in the reverse order to they wer popped.
To use a macro in you program you just use the name of the macro as an ordinary
macro instruction:
SaveRegs
;some other instructions
RestoreRegs
This example shows how you can use a macro to save typing in. This macro simply
prints out a variable to the screen.
The only problems with macros is that if you overuse them it leads to you program
getting bigger and bigger and that you have problems with multiple definition of labels
and variables. The correct way to solve this problem is to use the LOCAL directive for
declaring names inside macros.
Syntax:
LOCAL name
Where �name� is the name of a local variable or label.
If you have comments in a macro everytime you use that macro the comments will be
added again into your source code. This means that it will become unesescarily long.
The way to get round this is to define comments with a ;; instead of a ;. This example
illustrates this.
;a normal comment
;;a comment in a macro to save space
Macros with parameters:
Another useful property of macros is that they can have parameters. The number of
parameters is only restricted by the length of the line.
Syntax:
Name_of_Macro macro par1,par2,par3
;
;commands go here
endm
This is an example that adds the first and second parameters and puts the result in the
third:
AddMacro macro num1,num2,result
push ax ;save ax from being destroyed
mov ax,num1 ;put num1 into ax
add ax,num2 ;add num2 to it
mov result,ax ;move answer into result
pop ax ;restore ax
endm
On the next page there is an example of some a useful macro to exit to dos with a
specified . There are two versions of this program because both A86 and Turbo
Assembler handle macros differently.
;this is a simple program which does nothing but use a
;macro to exit to dos. This will only work with tasm and
;should be compiled like this: tasm /m2 macro.asm
;because tasm needs more than one pass to work out what
;to do with macros. It does cause a warning but this can
;be ignored.
.MODEL small
.STACK
.CODE ;start the code segment
Start: ;a good a place as any to start this
;now lets go back to dos with another macro
BackToDOS 0 ;the errorlevel will be 0
BackToDOS macro errorlevel
;;this is a macro to exit to dos with a specified
;;errorlevel given. No test is done to make sure that a
;;procedure is actually passed or it is within range.
mov ah,4Ch ;;terminate program and return
;;to DOS using
mov al,errorlevel ;;put errorlevel into al
int 21h ;;interrupt 21h function 4Ch
endm ;;end macro
end ;end program
Procedures that pass parametres:
Procedures can be made even more useful if they can be made to pass parameters to
each other. There are three ways of doing this and I will cover all three methods:
In registers,
In memory,
In stack.
In registers:
Advantages: Easy to do and fast.
Disadvantages: The is not many registers.
This is very easy to do, all you have to do is to is move the parameters into registers
before calling the procedure. This example adds two numbers together and then
divides by the third it then returns the answer in dx.
push ax ;save value of ax
push bx ;save value of bx
push cx ;save value of cx
mov ax,10 ;first parameter is 10
mov bx,20 ;second parameter is 20
mov cx,3 ;third parameter is 3
Call ChangeNumbers ;call procedure
pop cx ;restore cx
pop bx ;restore bx
pop ax ;restore dx
.....
ChangeNumbers PROC ;Defines start of procedure
add ax,bx ;adds number in bx to ax
div cx ;divides ax by cx
mov dx,ax ;return answer in dx
ret
ChangeNumbers ENDP ;defines end of procedure
How to use a debugger:
This is a good time to use a debugger to find out what your program is actually doing.
I am going to demonstrate how to use Turbo Debugger to check if this program is
working properly. First we need to compile this program to either a .EXE or .COM
file. Then type:
td name of file
Turbo Debugger then loads. You can see the instructions that make up your programs,
for example the first few lines of this program is shown as:
cs:0000 50 push ax
cs:0001 53 push bx
cs:0002 51 push cx
(This might be slighly different than is shown on your screen but hopefully you will get
the main idea)
This diagram shows what the Turbo Debuggerâ screen looks like
The numbers that are moved into the registers are different that the ones that we typed
in because they are represented in their hex form (base 16) as this is the easiest base to
convert to and from binary and that it is easier to understand than binary also.
At the left of this display there is a box showing the contents of the registers. At this
time all the main registers are empty. Now press F7 this means that the first line of the
program is run. As the first line pushed the ax register into the stack, you can see that
the stack pointer (SP) has changed. Press F7 until the line which contains mov
ax,000A is highlighted. Now press it again. Now if you look at the box which contains
the contents of the registers you can see that AX contains A. Press it again and BX
now contains 14, press it again and CX contains 3. Now if you press F7 again you can
see that AX now contains 1E which is A+14. Press it again and now AX contains A
again, A being 1E divided by 3 (30/3 = 10). Press F7 again and you will see that DX
now also contains A. Press it three more times and you can see that CX,BX and AX
are all set back to their origional values of zero.
Passing through memory:
Advantages: Easy to do and can use more parameters
Disadvantages: Can be slower
To pass parameters through memory all you need to do is copy them to a variable
which is stored in memory. You can use a varable in the same way that you can use a
register but commands with registers are a lot faster. This table shows the timing for
the MOV command with registers and then variables and then the amount of clock
cycles (the speed - smaller faster) it takes to do them.
KEY: reg8 means an 8 bit register eg AL
mem8 means an 8 bit variable declared with DB
reg16 means an 16 bit register eg AX
mem16 means an 16 bit variable declared with DW
imm8 means an immediate byte eg MOV al,8
imm16 means an immediate word eg MOV ax,8
Instruction
486
386
286
86
MOV reg/mem8,reg8
1
2/2
2/3
2/9
MOV reg,mem16,reg16
1
2/2
2/3
2/9
MOV reg8,reg/mem8
1
2/4
2/5
2/8
MOV reg16,reg/mem16
1
2/4
2/5
2/8
MOV reg8,imm8
1
2
2
4
MOV reg16,imm16
1
2
2
4
MOV reg/mem8,imm8
1
2/2
2/3
4/10
MOV reg/mem16,imm16
1
2/2
2/3
4/10
These timings are taking from the �Borlandâ Turbo Assemblerâ Quick Reference�
This shows that on the 8086 using variables in memory can make the instuction four
times as slow. This means that you should avoid using too many variables
unnecessarily. On the 486 it doesn�t matter as both instructions take the same amount
of time.
The method actually used is nearly identical to passing parameters in registers. This
example is just another version of the example given for passing in registers.
FirstParam db 0 ;these are all variables to store
SecondParam db 0 ;the parameters for the program
ThirdParam db 0
Answer db 0
.....
mov FirstParam,10 ;first parameter is 10
mov SecondParam,20 ;second parameter is 20
mov ThirdParam,3 ;third parameter is 3
Call ChangeNumbers
.....
ChangeNumbers PROC ;Defines start of procedure
push ax ;save ax
push bx ;save bx
mov ax,FirstParam ;copy FirstParam into ax
mov bx,SecondParam ;copy SecondParam into bx
add ax,bx ;adds number in bx to ax
mov bx,ThirdParam ;copy ThirdParam into bx
div bx ;divides ax by bx
mov Answer,ax ;return answer in Answer
pop bx ;restore bx
pop ax ;restore ax
ret
ChangeNumbers ENDP ;defines end of procedure
This way may seem more complicated but it is not really suited for small numbers of
this type of parameters. It is much more useful when dealing with strings or large
numbers of big values.
Passing through Stack:
This is the most powerful and flexible method of passing parameters. This example
shows the ChangeNumbers procedure that has been rewritten to pass its parameters
through the stack.
mov ax,10 ;first parameter is 10
mov bx,20 ;second parameter is 20
mov cx,3 ;third parameter is 3
push ax ;put first parameter on stack
push bx ;put second parameter on stack
push cx ;put third parameter on stack
Call ChangeNumbers
.....
ChangeNumbers PROC ;Defines start of procedure
push bp
mov bp,sp
mov ax,[bp+8] ;get the parameters from bp
mov bx,[bp+6] ;remember that first it is last out
mov cx,[bp+4] ;so number is larger
add ax,bx ;adds number in bx to ax
div cx ;divides ax by cx
mov dx,ax ;return answer in dx
ret
ChangeNumbers ENDP ;defines end of procedure
This diagram shows the contents of the stack for a program with two parameters:
To get a parameter from the stack all you need to do is work out where it is. The last
parameter is at BP+2 and then the next and BP+4.
Files and how to used them:
Files can be opened, read and written to. DOS has some ways of doing this which save
us the trouble of writing our own routines. Yes, more interrupts. Here is a list of
helpful functions of interrupt 21h that we are going to need to use for our simple file
viewer.
Function 3Dh: open file
Opens an existing file for reading, writing or appending on the specified drive and
filename.
INPUT:
AH = 3Dh
AL = access mode
bits 0-2 000 = read only
001 = write only
010 = read/write
bits 4-6 Sharing mode (DOS 3+)
000 = compatibility mode
001 = deny all (only current program can access file)
010 = deny write (other programs can only read it)
011 = deny read (other programs can only write to it)
100 = deny none
DS:DX = segment:offset of ASCIIZ pathname
OUTPUT:
CF = 0 function is succesful
AX = handle
CF = 1 error has occured
AX = error code
01h missing file sharing software
02h file not found
03h path not found or file does not exist
04h no handle available
05h access denied
0Ch access mode not permitted
What does ASCIIZ mean? An ASCIIZ string is a ASCII string with a zero on the end
(instead of a dollar sign).
Important: Remember to save the file handle it is needed for later.
How to save the file handle:
It is important to save the file handle because this is needed to do anything with the
file. Well how is this done? There are two methods, which is better?
Copy the file handle into another register and don't use that register.
Copy it to a variable in memory.
The disadvantages with the first method is that you will have to remember not to use
the register you saved it in and it wastes a register that can be used for something more
useful. We are going to use the second. This is how it is done:
FileHandle DW 0 ;use this for saving the file
;handle
MOV FileHandle,ax ;save the file handle (same as
;FileHandle=ax)
Function 3Eh: close file
Closes a file that has been opened.
INPUT:
AX = 3Eh
BX = file handle
OUTPUT:
CF = 0 function is sucsessful
AX = destroyed
CF = 1 function not sucsessful
AX = error code - 06h file not opened or unautorized handle.
Important: Don't call this function with a zero handle because that will close the
standard input (the keyboard) and you won't be able to enter anything.
Function 3Fh: read file/device
Reads bytes from a file or device to a buffer.
INPUT:
AH = 3Fh
BX = handle
CX = number of bytes to be read
DS:DX = segment:offset of a buffer
OUTPUT:
CF = 0 function is successful
AX = number of bytes read
CF = 1 an error has occured
05h access denied
06h illegal handle or file not opened
If CF = 0 and AX = 0 then the file pointer was already at the end of the file and no
more can be read. If CF = 0 and AX is smaller than CX then only part was read
because the end of the file was reached or an error occured.
This function can also be used to get input from the keyboard. Use a handle of 0, and it
stops reading after the first carriage return, or once a specified number of characters
have been read. This is a good and easy method to use to only let the user enter a
certain amount of characters.
Note: If you are using A86 this will cause an error. Change @data to data to make it
work.
.MODEL small
.STACK
.CODE
mov ax,@data ;base jaddress of data
mov ds,ax ;segment
lea dx,FileName ;put address of fileneame in dx
mov al,2 ;access mode - read and write
mov ah,3Dh ;function 3Dh -open a file
int 21h ;call DOS service
mov Handle,ax ;save file handle for later
jc ErrorOpening
mov dx,offset Buffer ;address of buffer in dx
mov bx,Handle ;handle in bx
mov cx,100 ;amount of bytes to be read
mov ah,3Fh ;function 3Fh - read from file
int 21h ;call dos service
jc ErrorReading
mov bx,Handle ;put file handle in bx
mov ah,3Eh ;function 3Eh - close a file
int 21h ;call dos service
mov di,OFFSET Buffer+101 ;Where to put the "$"
mov byte ptr [di],"$" ;Put it at es:di
mov ah,9 ;write a string to the
mov dx,OFFSET Buffer ;screen using function 9h
int 21h ;of interrupt 21h
mov AX,4C00h ;terminate program and return to
;DOS using
INT 21h ;interrupt 21h function 4CH
ErrorOpening:
mov dx,offset OpenError ;display an error
mov ah,09h ;using function 09h
int 21h ;call dos service
mov ax,4C01h ;end program with an errorlevel of 1
int 21h
ErrorReading:
mov dx,offset ReadError ;display an error
mov ah,09h ;using function 09h
int 21h ;call dos service
mov ax,4C02h ;end program with an errorlevel of 2
int 21h
.DATA
Handle DW ? ;variable to store file handle
FileName DB "C:\test.txt",0 ;file to be opened
OpenError DB "An error has occured(opening)!$"
ReadError DB "An error has occured(reading)!$"
Buffer DB 101 dup (?) ;buffer to store data from
;file one bigger for $
END
Function: 3Ch: Create file
Creates a new empty file on a specified drive with a specified pathname.
INPUT:
AH = 3Ch
CX = file attribute
bit 0 = 1 read-only file
bit 1 = 1 hidden file
bit 2 = 1 system file
bit 3 = 1 volume (ignored)
bit 4 = 1 reserved (0) - directory
bit 5 = 1 archive bit
bits 6-15 reserved (0)
DS:DX = segment:offset of ASCIIZ pathname
OUTPUT:
CF = 0 function is successuful
AX = handle
CF = 1 an error has occured
03h path not found
04h no availible handle
05h access denied
Important: If a file of the same name exists then it will be lost. Make sure that there
is no file of the same name. This can be done with the function below.
Function 4Eh: find first matching file
Searches for the first file that matches the filename given.
INPUT:
AH = 4Eh
CX = file attribute mask (bits can be combined)
bit 0 = 1 read only
bit 1 = 1 hidden
bit 2 = 1 system
bit 3 = 1 volume label
bit 4 = 1 directory
bit 5 = 1 archive
bit 6-15 reserved
DS:DX = segment:offset of ASCIIZ pathname
OUTPUT:
CF = 0 function is successful
[DTA] Disk Transfer Area = FindFirst data block
Example of checking if file exists:
File DB "C:\file.txt",0 ;name of file that we want
LEA dx,File ;address of filename
MOV cx,3Fh ;file mask 3Fh - any file
MOV ah,4Eh ;function 4Eh - find first file
INT 21h ;call dos service
JC NoFile
;print message saying file exists
NoFile:
;continue with creating file
This example program creates a file and then writes to it.
.MODELSMALL
.STACK
.CODE
mov ax,@data ;base jaddress of data
mov ds,ax ;segment
mov dx,offset StartMessage ;display the starting
;message
mov ah,09h ;using function 09h
int 21h ;call dos service
mov dx,offset FileName ;put offset of filename in dx
xor cx,cx ;clear cx - make ordinary file
mov ah,3Ch ;function 3Ch - create a file
int 21h ;call DOS service
jc Error ;jump if there is an error
mov dx,offset FileName ;put offset of filename in dx
mov al,2 ;access mode -read and write
mov ah,3Dh ;function 3Dh - open the file
int 21h ;call dos service
jc Error ;jump if there is an error
mov Handle,ax ;save value of handle
mov dx,offset WriteMe ;address of information to
;write to the file
mov bx,Handle ;file handle for file
mov cx,38 ;38 bytes to be written
mov ah,40h ;function 40h write to file
int 21h ;call dos service
jc error ;jump if there is an error
cmp ax,cx ;was all the data written? Does
;ax=cx?
jne error ;if it wasn't then there was an error
mov bx,Handle ;put file handle in bx
mov ah,3Eh ;function 3Eh - close a file
int 21h ;call dos service
mov dx,offset EndMessage ;display the final message
;on the screen
mov ah,09h ;using function 09h
int 21h ;call dos service
ReturnToDOS:
mov AX,4C00h ;terminate program and return to DOS
int 21h ;using interrupt 21h function 4CH
Error:
mov dx,offset ErrorMessage ;display an error message
;on the screen
mov ah,09h ;using function 09h
int 21h ;call dos service
jmp ReturnToDOS ;lets end this now
.DATA
StartMessage DB "This program creates a file called�,
�NEW.TXT in the C: directory.$"
EndMessage DB 0Ah,0Dh,"File create OK, look at�,
�file to be sure.$"
Handle DW ? ;variable to store file handle
ErrorMessage DB "An error has occurred!$"
WriteMe DB "HELLO, THIS IS A TEST, HAS IT�,
�WORKED?",0 ;ASCIIZ
FileName DB "C:\new.txt",0
END
How to find out the DOS version:
In many programs it is necessary to find out what the DOS version is. This could be
because you are using a DOS function that needs the revision to be over a certain
level.
Firstly this method simply finds out what the version is.
mov ah,30h ;function 30h - get MS-DOS version
int 21h ;call DOS function
This function returns the major version number in AL and the minor version number in
AH. For example if it was version 4.01, AL would be 4 and AH would be 01. The
problem is that if on DOS 5 and higher SETVER can change the version that is
returned. The way to get round this is to use this method.
mov ah,33h ;function 33h - actual DOS version
mov al,06h ;subfunction 06h
int 21h ;call interrupt 21h
This will only work on DOS version 5 and above so you need to check using the
former method. This will return the actual version of DOS even if SETVER has
changed the version. This returns the major version in BL and the minor version in BH.
Fast string print:
We have been using a DOS service, function 9 of interrupt 21h to print a string on the
screen. This isn�t too fast nor does it allow us to use different colours or position the
text. There is another way to print a string to the screen - direct to memory. This is
harder as you have to set up everything manually but it has a lot of benifits mainly
speed.
TextAttribute db 7 ;contains the character attribute
;default is grey on black
......
FastTextPrint PROC
.286 ;need this for shift instructions. Take out if
less
;than 286
;========================================================
;INPUT: AH - Row
; AL - Column
; CX - Length of string
; DS:DX - The string
; TextAttribute - the colour of the text
;OUTPUT: none
;========================================================
push ax bx cx dx bp si di es ;save registers
mov bl,ah ;move row into bl
xor bh,bh ;clear bh
shl bx,5 ;shift bx 5 places to the left
mov si,bx ;move bx into si
shl bx,2 ;shift bx 2 places to the left
add bx,si ;add si to bx
xor ah,ah ;clear ah
shl ax,1 ;shift ax 1 place to the left
add bx,ax ;add ax onto bx
mov ax,0b800h ;ax contains text video memory
mov es,ax ;move ax into es
mov si,dx ;mov dx into si
FastTextPrintLoop:
mov ah,ds:[si] ;put the char at ds[si] into ah
mov es:[bx],ah ;move the char in ah to es[bx]
inc si ;increment si (si+1)
inc bx ;increment bx (bx+1)
mov ah,TextAttribute ;put the attribute into ah
mov es:[bx],ah ;put ah into es position at bx
inc bx ;increment bx (bx+1)
loop FastTextPrintLoop ;loop CX times
pop es di si bp dx cx bx ax ;restore registers
ret ;return
FastTextPrint ENDP
Explanation of new terms in this procedure:
In this procedure there was several things that you have not come across before. Firsly
the lines:
push ax bx cx dx bp si di es ;save registers
pop es di si bp dx cx bx ax ;restore registers
This is just an easier way of pushing and popping more than one register. When TASM
(or A86) compiles these lines it translates it into separate pushes an pops. This way
just saves you time typing and makes it easier to understand.
Note: To make these lines compile in A86 you need to put commas (,) in between the
registers.
This line might cause difficulty to you at first but they are quite easy to understand.
mov ah,ds:[si] ;put the char at ds[si] into ah
What this does is to move the number stored in DS at the location stored in SI into
AH. It is easier to think of DS being like an array in this command. It is the same as
this line in C.
ah = ds[si];
Shifts:
There are four different ways of shifting numbers either left or right one binary
position.
SHL Unsigned multiple by two
SHR Unsigned devide by two
SAR Signed devide by two
SAL same as SHL
The syntax for all four is the same.
Syntax:
SHL operand1,operand2
Note: The 8086 cannot have the value of opperand2 other than 1. 286/386 cannot
have operand2 higher than 31.
Using shifts is a lot faster than using MUL�s and DIV�s.
Loops:
Using Loop is a better way of making a loop then using JMP�s. You place the amount
of times you want it to loop in the CX register and every time it reackes the loop
statement it decrements CX (CX-1) and then does a short jump to the label indicated.
A short jum means that it can only 128 bytes before or 127 bytes after the LOOP
instuction.
Syntax:
Loop Label
Using graphics in mode 13h:
Mode 13h is only availible on VGA, MCGA cards and above. The reason that I am
talking about this card is that it is very easy to use for graphics because of how the
memory is arranged.
First check that mode 13h is possible:
It would be polite to tell the user if his computer cannot support mode 13h instead of
just crashing his computer without warning. This is how it is done.
CheckMode13h:
;Returns: DX=0 Not supported, DX=1 supported
mov ax,1A00h ;Request video info for VGA
int 10h ;Get Display Combination Code
cmp al,1Ah ;Is VGA or MCGA present?
je Mode13hSupported ;mode 13h is supported
xor dx,dx ;mode 13h isn�t supported dx=0
Mode13hSupported:
mov dx,1 ;return mode13h supported
Just use this to check if mode 13h is supported at the beginning of your program to
make sure that you can go into that mode.
Note: I have not tested this on a computer that doesn�t hav VGA as I don�t have any.
In theory this should work but you should test this on computers that don�t have VGA
and see if it works this out.
Setting the Video Mode:
It is very simple to set the mode. This is how it is done.
mov ax,13h ;set mode 13h
int 10h ;call bios service
Once we are in mode 13h and have finished what we are doing we need to we need to
set it to the video mode that it was in previously. This is done in two stages. Firstly we
need to save the video mode and then reset it to that mode.
VideoMode db ?
....
mov ah,0Fh ;function 0Fh - get current mode
int 10h ;Bios video service call
mov VideoMode,al ;save current mode
;program code here
mov al,VideoMode ;set previous video mode
xor ah,ah ;clear ah - set mode
int 10h ;call bios service
mov ax,4C00h ;exit to dos
int 21h ;call dos function
Now that we can get into mode 13h lets do something. Firstly lets put some pixels on
the screen.
Function 0Ch - Write Graphics Pixel
Makes a color dot on the screen at the specified graphics coordinates.
INPUT:
AH = 0Ch
AL = Color of the dot
CX = Screen column (x coordinate)
DX = Screen row (y coordinate)
OUTPUT:
Nothing except pixel on screen.
Note: This function performes exclusive OR (XOR) with the new color value and the
current context of the pixel of bit 7 of AL is set.
mov ah,0Ch ;function 0Ch
mov al,7 ;color 7
mov cx,160 ;x position -160
mov dx,100 ;y position -100
int 10h ;call bios service
This example puts a pixel into the middle of the screen in a the color grey. The
problem with this method is that calling interrupts is really slow and should be avoided
in speed critical areas. With pixel plotting if you wanted to display a picture the size of
the screen you would have to call this procedure 64,000 times (320 x 200).
Some optimizations:
This method isn�t too fast and we could make it a lot faster. How? By writing direct to
video memory. This is done quite easily.
The VGA memory starts at 0A000h. To work out where each pixel goes you use this
simple formula:
Location = 0A000h + Xposition + (Yposition x 320)
Location is the memory location which we want to put the pixel.
This procedure is quite a fast way to put a pixel onto the screen. Thanks go to Denthor
of Asphyxia as I based this on his code.
PutPixel PROC
.286 ;enable 286 instructions for shifts remove if
;you have less than an 286.
;========================================================
;INPUT: BX=X postion
; DX=Y position
; CL=colour
;OUTPUT: None
;========================================================
;this can be optimized by not pushing ax if you don�t
;need to save it. For A86 change push ds es ax to push
;ds,es,ax and do the same thing with pop.
push ds es ax ;save ds,es and ax
mov ax,0A000h ;ax contains address of video
mov es,ax ;es contains address of video
mov di,bx ;move x position into di
mov bx,dx ;mov y postion into bx
shl dx,8 ;shift dx 8 places to the left
shl bx,6 ;shift bx 6 places to the left
add dx,bx ;add dx and bx together
add di,bx ;add di and bx together
mov al,cl ;put colour in al
stosb ;transfer to video memory
pop ax es ds ;restore ds,es and ax
ret
PutPixel ENDP
Thank you for reading. I hope that you have learnt something from this. If you
need any more help then email me.
Gavin Estey 19/2/95
Gavin�s Introduction to Assembly Page 1
[Back to TUTOR SWAG index] [Back to Main SWAG index] [Original]